Advanced Calculus Single Variable

4.7.3 Compactness And Open Coverings

In Proposition 4.7.12 it was shown that sequential compactness in ℝ is the same as closed and
bounded. Here we give the traditional definition of compactness and show that this is also
equivalent to closed and bounded.

Definition 4.7.16A set K is called compact if whenever C is acollection ofopen sets such that K ⊆∪C, there exists a finite subset of open sets

{U1,⋅⋅⋅,Um }

⊆Csuch that K ⊆∪i=1mUi. In words, it says that every open cover admits a finite subcover.

Proposition 4.7.17Every closedinterval

[a,b]

is compact.

Proof: Suppose not. Then there exists an open cover C which admits no finite subcover.
Consider the two intervals

[a, a+b]
2

,

[a+b,b]
2

. At least one of these fails to have a finite open
cover from C. Otherwise, there would be a finite open cover of

[a,b]

. Pick the interval which
fails to have a finite open cover from C. Call this interval I2. It is half the length of
I1≡

[a,b]

. Now do for I2 exactly what was done for I1. Split it in half and take the
half which fails to have a finite subcover. Continue obtaining a nested sequence of
closed intervals such that the length of In is 2n−1 times the length of I1. By the
nested interval lemma, there exists a point p which is in all these intervals. Thus
p ∈ U ∈C. Therefore, there exists δ > 0 such that

(p − δ,p + δ)

⊆ U. Now for all n large
enough, the length of In which contains p, is less than δ. Hence, In is contained in U
contrary to the definition of In which required that it admit no finite subcover from C.
■

Now here is the main result, often called the Heine Borel theorem.

Theorem 4.7.18Let K be a nonempty set in ℝ. Then K is compact if andonly if K is closed and bounded.

. It follows that there are finitely many of the
open sets in D which cover K. However, ℝ ∖ K contains no points of K and so if this
finite set includes this one, you can simply delete it and still have an open cover of
K.

Conversely, suppose that K is compact. Why is it closed and bounded? Suppose first it is
not closed. Then there exists a limit point p which is not in K. By Theorem 4.7.8 there exists
a sequence of distinct points of K

{pn}

such that limn→∞pn = p

∕∈

K. Then p is the only
limit point of this sequence. Hence

∞
Cn ≡ ∪k=npk ∪ {p}

is a closed set. Then since p

∕∈

K,

{ℝ ∖ Cn}∞n=1

is an open cover of K but it admits no finite subcover. This is because the sequence of open
sets is increasing and ℝ ∖ Cn fails to contain pn∈ K. Thus K is closed. Why is K
bounded? If not, there exists a sequence

{kn}

n=1∞⊆ K such that

|kn|

> n. This
sequence cannot have any limit points. Hence ∪j=n∞kj≡ Cn is a closed set. The
open sets

{ℝ ∖Cn }

n=1∞ provide an open cover which admits no finite subcover
since it is an increasing sequence of open sets, the nth of which fails to include kn.
■

This theorem and the earlier result shows that in ℝ, sequential compactness, compactness
and closed and bounded are all the same thing. The same conclusion can be drawn for
ℂ.